 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Grape Consumption Conclusion (Posted on 2016-03-08) A group of children held a grape-eating contest.
When the contest was over, the winner had eaten N grapes, where N is a positive integer.
The child in Mth place had eaten N + 2 – 2M grapes.
The total number of grapes eaten in the contest was 2016.

Find the smallest possible value of N.

 No Solution Yet Submitted by K Sengupta Rating: 2.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution with no ties Comment 4 of 4 | I am solving this problem with the provision that each child ate a different number of grapes.

The number of grapes eaten by each child's place forms an arithmetic sequence: N, N-2, N-4, ... N-2(M-1), assuming there are M children.  The sum of the sequence is M*(N + N-2*(M-1))/2 = M*(N-M+1).

Then M*(N-M+1) = 2016 implies N = 2016/M + M - 1.  For N to be an integer, M must be a factor of 2016.  Subsequently the sum 2016/M + M is the sum of a two term factorization of 2016.

In a continuous system the minimum value occurs when M = sqrt(2016) = 44.899889.  In the integers, the closest we can get is 2016 = 42*48.  Then N = 42 + 48 - 1 = 89.

This checks out: with 42 children and the first place child eating 89 grapes, the last place child eats 7 grapes and the sum of the series 89+87+...+7 = 2016.

 Posted by Brian Smith on 2016-03-08 10:32:34 Please log in:
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